text:
Currency - is a safety pilot a passenger per the regs (FAA)?
Per the regs in FAR 61.57, for a pilot trying to get current (for night or day) that is presently not current for carrying passengers, in a Cessna 172 where only one crew member is required, would a properly rated safety pilot sitting in the right seat but not acting as PIC count as a passenger? The regs don't seem so clear in this matter. It's always good to have a safety pilot if you haven't done something in a while, especially at night.

You mention "Safety Pilot", are you specifically talking IFR currency? You mention day/night which makes me think VFR, but typically you don't use safety pilots for VFR operations.

@RonBeyer I'm just using the term safety pilot generically - a second set of eyes to keep you from making an oops at night. Always good to have two people looking at the altimeter and airspeed on final at night, especially if one were out of practice :-)

Safety Pilots are usually there to maintain visual separation with traffic, weather, and terrain, and can log time either as PIC or SIC for IFR currency. If you just have a pilot rated friend tag along for some extra eyes, they are a passenger, unless they are acting as PIC. See the answer I linked (and the Wikipedia article) for more info.

How should I deal with this question if I find the link you sent satisfactory?

Its not negatively impacting your reputation, so you can leave it and let it drop off, that will make it better for somebody else who comes along and searches for wording similar to yours.


text:
What are the rules for PIC change during flight?
Per the FAA FARs - Are there any rules governing a "change of who is the PIC during flight, and thus who can log hours? Say a friend and I go flying together - can we switch as PIC half way through and each log half the time? Also, would the time not spent as PIC (if it is possible to switch) be logged as total flight time in addition to the time spent as PIC?

Acting as PIC and logging PIC time are two different things. I guess you're really more interested in how to log PIC time?

I think you are confusing the terms "pilot in command" and "pilot flying". They are not the same thing.

@J.Hougaard I mean PIC per the definition of sole manipulator of the controls as defined in part 1. Does that change anything?

@Pugz the sole manipulator of the controls is not necessarily the pilot in command

@J.Hougaard I understand that. I suppose I'm just trying to get to the root of who can log PIC time if they are the "acting PIC"

I'm not really sure what you're asking. Do you want to know how to change the acting PIC, or how to change who's logging PIC? Or, as per your last comment, are you asking "when can the acting PIC also log PIC time?". You might also consider asking about this in chat if you - or we - need help to express things unambiguously.

@ymb1 is correct in that it does not matter who the legal pilot in command is per §1.1.  What does matter is following the rules in §61.51(e).

(e) Logging pilot-in-command flight time. (1) A sport, recreational, private, commercial, or airline transport pilot may log pilot in command flight time for flights-

(i) When the pilot is the sole manipulator of the controls of an aircraft for which the pilot is rated, or has sport pilot privileges for that category and class of aircraft, if the aircraft class rating is appropriate;
(ii) When the pilot is the sole occupant in the aircraft;
(iii) When the pilot, except for a holder of a sport or recreational pilot certificate, acts as pilot in command of an aircraft for which more than one pilot is required under the type certification of the aircraft or the regulations under which the flight is conducted; or
(iv) When the pilot performs the duties of pilot in command while under the supervision of a qualified pilot in command provided—

If in your example both pilots are rated to fly that airplane on your license whether or not you have the required endorsements or ratings to act as the legal pilot in command you can log PIC time.  It does not matter who the legal pilot in command is at that time.  Both pilots will log the flight time as PIC and total time for the time you are the sole manipulator of the controls.

Other ways both pilots can log PIC time is where one pilot who is the safety pilot is the legal pilot in command while the other pilot who is flying the airplane and sole manipulator of controls is under simulated instrument flight.
Let me be very clear here.  If both pilots were to log PIC time under this condition, and many pilots do, they are making the statement in their logbook that could get them in trouble later on.  Let me explain, pilot A and B go out and fly together; Pilot A is under the hood and the other is the legal pilot in command.  On another day, Pilot B is under the hood and the other is the legal pilot in command.  Both pilots log exactly the same amount of flight time for both PIC and total time on these flights.  At this point, a precedence is set.
What is that statement you are making in your logbook?
Who the legal pilot in command is for each flight.  Read the regulation again.  If the safety pilot logs PIC time they are accepting legal pilot in command responsibilities and all that goes with it.  If the pilot under the hood has an incident or accident or airspace infraction the FAA will want to know who the legal pilot in command was during that flight.  The logbooks tell the story regardless what the pilots say at that moment.

That is the correct answers. The OP is confusing acting as PIC with logging PIC which are not the same. If you're sole manipulator of the controls of an aircraft you are rated for you log PIC. Doesn't make any difference who the PIC is.

Just to clarify part one of your answer. In the OPs question the plane does not require two pilots, so only the person manipulating the controls can log PIC time. Total logged time for both pilots would be equal to the total time of the flight.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
How is training for an SIC type rating accomplished when an aircraft is two pilot certified only?
For an aircraft which is certified for two pilots and a simulator doesn’t exist, how does someone go about receiving the required training for an SIC type rating without having the type rating already, since two type certificates pilots are required for the operation of the aircraft? Furthermore, how can the pilot in SIC training log anything to meet the experience requirements for the SIC type rating when you cannot log anything for a type in which you are not rated (category, class, and type), much less act as a required crew member? This question is specifically in regard to FAA regulations.

Last I heard, an SIC type is the result of some sort of international standardization.  There is no requirement to be "typed" as an SIC to serve as an SIC while flying a US registered aircraft within the NAS.  As such, no training is required to occupy the SIC position other than to be category and class rated.  So, you and the PIC can legally fly the plane while you receive your SIC type training.  Now insurance coverage is another story.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How do symmetrical airfoils generate lift?
It is widely understood that the curved shape of a typical airfoil (such as a Clark Y) causes a pressure difference that creates lift. How do symmetrical airfoils do this when the top and lower surfaces are the same shape and length?

Actually it is widely understood that the curved shape is not needed to create lift. Flat plate at slight angle to the flow does generate lift. Not as efficiently, but using the same principle. Also, the length is irrelevant, because the air flowing over the upper surface will cross the wing in significantly shorter time than the air flowing over the lower surface.

See also How do paper airplanes create lift if their wings are flat? and What is the performance of a flat plate wing? for explanation why even flat plate generates lift.

Even the Wikipedia article on Lift calls out the common misconception of equal transit times as completely unfounded in reality.

By using a non-zero angle of attack.  When the trailing edge is pointed downwards, and assuming the airstream leaves the trailing edge smoothly, the exiting airstream is deflected downwards.  This causes lift via conservation of momentum.  Increasing the angle of attack will increase your lift until such time as the airstream over the trailing edge becomes non-smooth.  (When this happens, you're close to stalling.)

(Image from this page, which unfortunately appears to be down)
As you can see from the above graph, a symmetric airfoil at zero angle of attack generates no lift;  see this site from NASA as well as the above Wikipedia page.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
What are the structures on the sides of and aft of the canopy on Israeli F-16s?
I know Israel has highly modified F-16s, but just what are the structures on the sides and top, aft of the canopy? See this image of an Israeli F-16 compared to a normal one:

Vs

Even though this has already been marked duplicate, please properly  attribute your images to their owners.

They're conformal fuel tanks, a type of external fuel tank designed for a specific aircraft type to reduce drag and other effects.
In general, they increase fuel load and free up stations for weapons or even more fuel.
In particular, the F-16 is a small platform with a modest fuel load and thus has a moderate radius even when clean. But when it's combat loaded with sensors and weapons, the additional weight and drag quickly eat into its combat radius, so one or two drop tanks are semi-permanently hung from its wings and/or belly. Don't quote me, but the CFT's almost double the F-16's internal fuel capacity.
The F-16 wasn't originally conceived for ground attack, carrying bombs for many of its missions, but it quickly grew into the versatile multirole platform it is today.
The USAF can better accommodate the F-16's short range because it flies a huge fleet of ~450 aerial tankers to service 2000 fighters, whereas the IAF operates just 11 tankers for 400 fighters. (Obviously other USAF a/c also require tankers, not just fighters.) Also, the IAF doesn't have the luxury of other nearby friendly air bases it can fly out of. Thus, they use CFT's.
With the benefit of hindsight, the original F-16 design team would probably have enlarged the internal fuel tanks from the start had they known how the F-16 would be used.

Those are conformal fuel tanks

While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference.  Link-only answers can become invalid if the linked page changes. - From Review

@federico Actually, the essential part of the answer is included here. Strip out the link and the answer remains, but hey, what do I know?

To me 5 words do not make a complete answer. See for example the accepted answer to the original question (of which this is a duplicate)

It answered the question I asked ;-)


text:
Can you log PIC for dual time after you have your PPL?
I've looked up a few different regulations (FAA) regarding the logging of time in my logbook and am unclear on how to log my time.
Here's the scenario: I am a private pilot and am getting checked out on a new aircraft. I am sitting in the left seat and am legal to operate the aircraft currency/certificate-wise, but a CFI is taking me through the motions with a new plane. How should I log this? Dual, PIC, or both?
Also, here's scenario two: everything is same as in the above scenario except I am getting checked out on a complex aircraft for my complex endorsement. How should I log this?

I saw this and was still not 100% clear... the interpretation offered by Lnafziger says "you could log both PIC and dual..." but nowhere in the regs is anything stated about both logging both in the same place. Just trying to cover myself.

You could also look here - http://aviation.stackexchange.com/questions/12735/must-i-log-aircraft-checkout-flights-as-dual-received?rq=1 - there's so much confusion amongst pilots about logging, and our questions reflect that, sadly.

In both situations you will log both DUAL and PIC.


text:
Why is there a GPS/RNAV waypoint co-located with a VOR?
There is an RNAV waypoint named STYCK co-located at what appears to be the exact location of the LOA VOR (coordinates match precisely as well). Why would the creation of a waypoint be necessary if IFR GPS/RNAV units/databases can take both VOR names and waypoints? Why not just use the VOR name that has existed there for over 50 years? Such a thing is regularly done in IFR SIDs and STARs when loading them into a GPS/FMS.

In the US, the FAA has done this alot, as in their long term plan to phase out a lot of the VORs. Doing that, they're preemptively just adding a Fix for procedures on top of the VOR so they don't have to update all the procedures when the VOR is decommissioned.

Agreed. Any reference?

No references, just discussion with the people who go around designing the RNAV arrivals, when we were designing some for LIT.

phase out a lot of the VORs - in that case, they'd have to phase out a lot of the small Cessna aircraft also - or what are those going to navigate with ? XD

Programmer, there aren't a lot of non-GPS Cessnas out there, anymore. GPS is cheap and fits into a small space. Pilots of such aircraft usually take a hand-held GPS along, although these units are not certified for flying approaches.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
What are these parts on STOL/bush planes?
I've seen these (note the parts pointed to by the red arrows in the image) on the landing gear struts of Super Cubs and plenty of other STOL bush planes with large tundra tires. It looks like maybe just an aerodynamic fairing on the gas tube/strut of the landing gear but I'm wondering why on every example I've seen of this, the aerodynamic shapes are angled differently (one is more "up" than the other). What is this if I'm wrong about what it is and most of all why are they always angled differently? Is this to try and help with adverse yaw or something?
You can also see them in this video here (note that they, too, like all I've seen, are angled differently): FlightChops Super Cub

You say they are always angled differently.  Is this difference always consistent with the direction of the spiral of the slipstream?  In this photo, it is.  If the trend is consistent, someone might want to include that observation in an answer.  An answer has already noted that they are loose fitting and free to move slightly.

Those are not specifically STOL hardware, they are bungee covers that happen to have an airfoil shape.
Aircraft like the one in your picture use bungees (also called "shock cords") to absorb the force of landing: As weight is transferred to the gear the bungees are stretched. The installed bungees look like this:

The bungees are literally just "aviation-grade" bungee cords, and would be quickly deteriorated by weather if left exposed, so they are usually covered. The most basic covering is a leather or plastic boot:

The boots are not much more aerodynamic than the exposed bungees however, so some aircraft (including many with STOL kits) have a cover with an aerodynamically favorable shape to reduce the drag caused by the bungee bulge. That style of cover is what's shown in your picture.
Here's a closer look:

Any idea why each of the aerodynamic covers are ALWAYS at a different angle from the other?

@Pugz they aren't always at a different angle from each other - in fact they should be at the same angle (so the "fat" end of the fairing is aligned with the relative wind in cruise flight - splitting the airflow around the bulge while creating the least amount of drag). They may appear to be at different angles in photographs (particularly those taken at an angle) or be slightly offset on the ground if the gear is uneven though. If I could find a good head-on eye-level shot of an installed pair it would be a better illustration but that's not a popular photographing angle.

The covers are a loose fit over the bungee and strut and are free to rotate a fraction of a turn.

Those have wing airfoils, so it has to do something with increasing lift. Haven't seen you Pugz in about a month.

;-) It feels so good to be baaaack!!!

@Pugz So far none of the ones on this page look like they're at a different angle from each other, including the picture you posted. It is very likely that you are seeing an optical illusion where to me they look like they're aligned correctly.


text:
How does the PTT button on the radio work?
I've always seen a button either built into the yoke or attached to the yoke with velcro on older aircraft. However, the other day my instructor used an older-looking headset that had a push to talk button built into the side of the headphones. He had no cable plugged in except the mic/headphone jacks and never touched the yoke's PTT button to transmit. My question is: how is the PTT function activated? Is this done with a load resistor or something that loads down the mic line, triggering radio transmission? I've done some research and looked at various manuals of audio panels and navcoms and haven't found an answer. The audio panel in use in this case is the Garmin GMA-430 audio panel, pictured here:

This page has a pretty good explanation of the basics.  In essence, the PTT switch operates to circuits or relays.  One to transmit and one to mute the receiver.

I'm not sure what you are asking, it seems you wonder how the commutation between transmission and reception is done when the PTT is not used. If this is the case, then the answer is a vox circuit is used to detect audio from the mike and perform the PTT-equivalent function when the sound from the mike exceeds some adjustable level. Here is a headset with a vox included. The sensitivity button is visible.

The standard microphone plug on GA planes is the PJ-068 plug, which has three electrical contacts (tip, ring, and sleeve). The tip of this connector is a push-to-talk signal (the ring carries the actual audio signal, the sleeve is ground).

PTT on aircraft radios and audio panels is activated by taking the PTT contact (tip of the microphone plug) to ground or microphone low. All you're doing is literally closing a switch that grounds PTT and that switch can be located just about anywhere. Putting it on the headset makes it nearly impervious to damage from cable strain. It also makes it so he ALWAYS knows where the PTT button is, regardless of which aircraft he's in.


text:
Why do helicopters have windows near the pedals?
What is the purpose of the clear windows in some helicopters in the floor of the nose, right where the pilot's feet and pedals are?
For example, here's a Huey with windows clearly visible:

Airbus helicopter example:

I was involved in a Search & Rescue many years ago, using a C-130 Hercules. While 2 "spotters" were based at the side doors to the rear, there were 2 more looking through windows near the pilot / co-pilot feet. So the ability to look down is not limited to helicopters.

The usual purpose of windows is ....... to see through!

Honestly, I was tempted to downvote the question because the answer seemed so obvious. But for all I know, there could have been another reason well-known to pilots but not to civilians like Sam and me. My first question on this forum probably got a few actual facepalms.

These fall into a category known as vertical reference windows. They can be used for many purposes, but essentially, they allow us to gain a better spatial picture of the orientation of our aircraft, the skids, and anything that might be slung beneath the aircraft. As someone who has landed many helicopters on trailers, it is extremely helpful to be able to see through the floor when precisely placing the skids. Holding a pinnacle approach and hover is much easier to execute as well. Further, when you have something on a line, it makes it easier to keep it in view, especially when it is closer to the aircraft.

Also because when the engine quits and you start gliding, what you can see between your feet is where you're going lol.

@JohnK: Not necessarily. You will probably be carrying a good amount of forward airspeed during autorotation (up to 100 kts in a Bell 206B), so you’ll be looking well-forward of the aircraft.

I guess a better question might be why do (most) fixed-wing aircraft not have this feature?  It might be helpful to see the ground directly below you when landing a regular plane as well.

@DarrelHoffman: I guess because in most light, single, engine planes there’s an engine in front of you. Even in some light twins, they put other things up front like luggage or a cabin heater, etc. Plus, airplanes are usually still moving fairly quickly when they touch down. You want pilots to keep their eyes up, not looking down through the floor.

@AaronHolmes Yeah, if you're looking down at your feet during touchdown on an airplane, you're probably doing it wrong. Maintaining the centerline is much easier if you're looking at the runway way out in front of you, not directly underneath you. Now, for taxiing very long aircraft, being able to see the location of the nose gear relative to the pavement actually can be helpful, which is why some newer heavy jets have foward-facing cameras on the bottom of the fuselage aft of the nose gear. As a bonus, airlines can also choose to attach it to the IFE system and give the pax a cool view.

@DarrelHoffman Helicopters can hover and descend vertically, so floor windows can help you to look down and see where you're going. In general, if a fixed-wing aircraft is flying at an angle where floor windows would show you where you're going, something has gone very wrong.

@DarrelHoffman In a fixed-wing plane, by the time a particular patch of ground is directly below you, it's irrelevant -- it's going to be be behind you very shortly.

@anaximander: You're absolutely correct. I just want to point out, for accuracy's sake, that while helis CAN descend straight down, they usually maintain some forward momentum anyway, to avoid flirting with a settling-with-power situation.

Modern planes like the 747-8 have actually camera at the bottom to allow a view at the ground. The in-flight-entertainment usually allows a passenger to use it also. A front sight and a behind view (ruder) is usually also available.

They'd be damned handy in a taildragger. I wonder whether it would be possible with cameras and electronics? :)

@PeterWone Build an F35 HUD into your bush plane

@DKNguyen After I work out how to fit the missile launchers. Dang wabbit.

@PeterWone Either you're a really terrible hunter or you're hunting terribly dangerous rabbits.

@DKNguyen Australia, right? 6' tall with fangs. Similar to drop bears.

I assume because being able to see straight down helps in an aircraft capable of vertical movement and landing, especially when flaring. That would suck to only be able to see up when you're moving down towards a solid object.


text:
What are the differences between "Off/On Time", "Total AC Time", and "Offset"?
Can somebody please help explain to me the difference between all of these. I understand that "Off/On time" is probably the Hobbs time which is determined by the time my battery master switch is turned on - but I do not understand the total AC time and the Offset. I am assuming that one of these is the time that my plane is in the air. Picture attached for reference, and I fly a Cessna 172R.

There are different definitions for times in the U.S. FARs.
For example FAR §1.1 has a definition for flight time

Flight time means:

(1) Pilot time that commences when an aircraft moves under its own power for the purpose of flight and ends when the aircraft comes to rest after landing; or
(2) For a glider without self-launch capability, pilot time that commences when the glider is towed for the purpose of flight and ends when the glider comes to rest after landing.

This would typically be your OUT and IN time.
The Hobbs meter (depending on how it is wired) indicates the hours the engine is running or the hours the airplane is flying.
FAR §1.1

Time in service, with respect to maintenance time records, means the time from the moment an aircraft leaves the surface of the earth until it touches it at the next point of landing.

FAR §91.409(b)

Except as provided in paragraph (c) of this section, no person may operate an aircraft carrying any person (other than a crewmember) for hire, and no person may give flight instruction for hire in an aircraft which that person provides, unless within the preceding 100 hours of time in service the aircraft has received an annual or 100-hour inspection and been approved for return to service in accordance with part 43 of this chapter...

So the OFF-ON times are used for maintenance intervals.
I am personally unaware of total aircraft time or offset.
Most FMS installations have the ability to capture this data automatically.  It is called the OOOI value or OUT-OFF-ON-IN times.  Use the OFF-ON for maintenance and the OUT-IN for pilot logbooks.

I suspect 'offset' is related to replacing the meter. The meter was replaced at 3389 hours, so the aircraft's next service is due at 10617 hours, at which point the meter will read (10617-3389) = 7228 hours.

A Hobbs meter that is started when you turn on the battery master is probably the least accurate method of determining Air Time or Flight Time.
The pilot in command is responsible to determine the correct air time and flight time for each flight they operate.
Aircraft rental companies, and individual aircraft owners, often make up there own procedures for recording air time and flight time that are not necessarily accurate or legal.
The FAA Air Regs make no mention of any sort of hour meter. The best source for recording air time or flight time is the clock in the aircraft or the watch on your wrist.
“HOBBS” time has no legal status. This is especially true since there is no standard on how an hour meter is activated. Some hour meters start when the battery master is turned on. Some start when they sense engine oil pressure. Some start when only when the aircraft is airborne.
“TACH” time also really has no legal standing as it is RARLEY accurate for flight time or air time.
The fact that Hobbs or Tach time are routinely used in pilot and aircraft log books is just an example of convenience and laziness.
“Off/On Time” represents the actual time on a clock at the moment takeoff and landing, and used to calculate the actual air time for a particular flight.
“Total aircraft time” represents the total accumulated Air Time or an aircraft.
“Offset” is just an attempt to try and correct for the inaccuracies of using hour meters to record air time or flight time.

There are some interesting distinctions made here.  I'd be interested to know if they are traceable to, in order of preference: FAR's,, any FAA or ICAO publication, an approved set of opspecs.


text:
What are some of your favorite flying and piloting related lessons or concepts to share with your non-pilot friends?
My team at work is doing this thing where each person teaches something new every week. I want to teach them about flying (I'm a private pilot). I'm curious what you think are topics that people find interesting.
I want to skip explaining the physics/aerodynamics but instead focus on some lessons or concepts that can be transferred to other things in life.
For e.g

how everything is designed with safety / redundancy in mind,
single pilot resource management
checklists

What are some other cool ideas?

While interesting, your question is opinion based and therefore off topic.

Hi, welcome to aviation. While this question is not on topic on the main site you might well find some people interested in this in chat

There exists a machine with which you can casually swim through the atmosphere to explore mountain tops, deep canyons, and vast expanses.  You need no special degree or unusual intellect to use it.  In fact, morons do it!  A little over 100 years ago it would have been called magic.  Today they are so common that many consider them a nuisance or necessary inconvience.
One type (there are A LOT of different types) is a million pound machine that can carry 500 people seven miles above the earth at 85% the speed of sound and cross entire oceans in mere hours.  How do you just not fall to your knees, mouth agape, in awe of such a thing??
Oh...because there are so many of them, and they're so safe, and so cheap to use that anyone with a few hundred dollars can bum a ride to anywhere on the planet while garbed in clothing a notch above underwear.
The fact is that we are living in SciFi land from the perspective of someone three generations ago.
And that's just airplanes.


text:
Is 1,000ft vertical separation in a sparse airspace increasing risk unnecessarily?
On Saturday the 8th of August 2020 at 19:40:10 UTC, two flights flying in opposite directions, in the same corridor over India were vertically separated by just 1,000ft. The flights were SQ322 from Singapore to London at FL340 and CV7306 from Luxembourg to Singapore at FL330.
While this is allowed under ICAO Reduced Vertical Separation Minima, given the handful of airplanes in the sky due to the pandemic, is allowing just 1,000ft taking unnecessary risks?

This has been done all over the civilized world for years & years. What gives you reason to think that 1,000' of vertical separation is an "unnecessary risk?"  How much separation would no longer be in that category?

There have been several recent incidents where airplanes came so close that pilots had to rely on TCAS to avoid a collision. Accidents never happen due to a single error. At the legal minimum the planes are potentially less than 20 seconds apart should something go amiss. Given that there were no other planes in the sky, it could be argued that choosing the legal minimum may increase the (albeit very low) risk of a collision.

@alexth, TCAS activations are somewhat common, but it is always because the controller made a mistake, the pilots made a mistake, or because the one aircraft is told to level off 1,000 ft below (climbing) or above (descending) another and the vertical speed is more than 1,000 ft/min, so TCAS triggers before they reach it, because it is time-based. I've never heard of one when the levels were assigned correctly, but the aircraft failed to maintain it.

You should define why you think this is an "unnecessary risk" because as this question is currently worded it feels sort of opinion based. Some might consider going in a pressurised tube into an unbreathable altitude an unnecessary risk.

@JanHudec, thanks. Both the 1996 Charkhi Dadri and 2002 Überlingen mid-air collisions are examples where the planned vertical separation of 1,000ft was potentially a contributing factor in that it left little room for error. Neither ATC nor the pilots had enough time to notice and correct their mistakes.

@Jamiec, apologies, I am not working in the airspace industry. My background is in risk. By "unnecessary" I meant the cost of keeping the planes further apart was minimal in this instance.

So are you asking about the risk as in financial risk, or safety risk?

@Jamiec Safety risk. By cost I meant the additional effort, I assume from ATC, to reassign one of the flights to a slightly different flight level.

@alexth, both the 1996 Charkhi Dadri and 2002 Überlingen mid-air collisions are firmly in the territory of humans selecting the wrong altitude. In the later, they were assigned the same altitude, so different stepping would have no effect at all. In the former they busted the cleared level, so larger stepping might have helped, but it was still failure of humans to observe the correct altitude, not inability to follow it accurately enough.

@JanHudec that makes sense. Just wanted to reiterate that my question was in the context of an almost empty airspace. Thanks again.

@alexth, the thing is that vertically it's pretty crowded, because jet aircraft are most efficient between FL320 and FL400 depending on weight. Normally even levels are assigned to one direction, usually westbound, and odd levels the other direction, usually eastbound, and there are really few choices left.

@JanHudec That's very helpful, thank you.

@alexth I understand the thoughts that the airspace is empty all the time now, but this is just like 24/7 night operations. The aviation community gets safety through easy to apply consistent procedures that are designed with safety in mind and that apply no matter if there are 2 aircraft or 100 in the airspace. Though it can feel like it less safe, changing the procedures can result in increased risk in the miss application of the procedure.

Under usual conditions aircraft pass directly above each other at 1,000 ft vertical separation thousands of times a day and it's never been a problem¹. The autoflight systems can maintain altitude to 100 ft reliably, so 1,000 ft is more than enough (hand-flown aircraft must be separated by 2,000 ft from FL290 (29,000 ft) up).
The risk does not become higher because there are fewer aircraft flying around, so there is no reason to change the procedures. Quite contrary, changing procedures would carry an increased risk of human error, so it would be counter-productive.
And in this case, procedure include the typically assigned altitudes for route and direction. Best put the planes to the flight levels the pilots are used to.

¹ All mid-air collisions (under instrument rules) occurred because some of the humans involved selected incorrect altitude, never because the plane would not be maintaining the altitude.


text:
Why is the assumed Lift-to-Weight-Ratio for the limit drop test for heavy aircraft (FAR-25) higher than for small aircraft (FAR-23)?
The FAR-25.725 assumes a Lift-to-Weight-Ratio L/W = 1 for the landing gear drop test while the FAR-23.725 assumes only L/W=0.667. This factor affects the load inertia factor n which is also needed for structural analysis. Does anybody have an explanation for the lower assumed Lift-to-Weight-Ratio for smaller aircraft?

It's most likely one of these things that make Part 23 easier to certify than Part 25 (because the intended product carries fewer people). Plenty of these examples in S&C/perf/loads.

Thanks for your comment. I totally agree with it. However, I think there is also a physical explanation, why heavier aircraft have a lower load inertia factor. Even if you stay in the range of the FAR-23 design code it seems common engineer knowledge that the lift-to-weight-ratio increases (and the inertia load factor decreases) with the weight. Even after long research, I couldn't find any explanation.

Ah ok, in this case it would appear to be opposite. I believe it has something to do with Part 25 operators are professionals, whereas most Part 23 are general community (commuter category aside). It's the same thing with stall demo: Part 23 requires pulling all the way to full stop (or uncontrollable ND), whereas Part 25 only requires going till distinctive ND occurs. But this is slightly different in that this is specified through AC, not the reg.


text:
How much lift comes from the fuselage on modern jets?
I understand this question is broad, but specifically referencing modern jets where composites are used a lot to get complex surfaces (such as in the Beechcraft Premier 1), how much of the total lift is generated by the fuselage versus the wings, perhaps as a percentage of total lift? The fuselage is obviously a lifting body and I’ve noticed the same in other current designs.

This question is too general and there are too many unanswered variables to answer this question.

I would be happy with a “generally from x% to x% if such a figure is known by any designers around here.

A close approximation is to take the  wing lift using the general lift formula (CL for the airfoil, wing area, angle of attack). Subtract wing lift from aircraft weight, equals amount fuselage is lifting.  i.e. if aircraft weighs 10,000lbs and CL wing lift = 9500lb, then everything else (mostly fuselage - but some engine) is 500lbs.

I have run the numbers for small aircraft like C150 or C172 and my recollection was the fuselage and engine provided about 5-7% lift. However in aerodynamics school we ran numbers for the T33 and I think the fuselage could provide all the necessary lift at about 350kts - so speed matters. A B727 doubles it's wing area in the landing configuration, showing the wings on an airliner are really only needed for landing and takeoff!

@jwzumwalt, wings are needed for producing lift efficiently. Lifting bodies only make sense at hypersonic speeds.

Not as much as potentially possible. This thread lists the numbers for some common aircraft and should give you a general idea, it seems to be ~10%

747:

Wings: 89.5%
Fuselage: 13.1%
Horizontal Tail: -3.7%
Vertical Tail: .1%
Nacelles and Pylons: 1%

P-51D:

Wings: 92.9%
Fuselage: 7.7%
Horizontal Tail: -.7%
Vertical Tail: .1%

I belive the center bulge on the bottom is also there to allow for a floor recess in the cabin making it easier to get around.

It's not as much as potentially possible, because it is a really bad idea. You only want to create fraction of lift that a little over the fraction of its width to wingspan. Producing more lift with fuselage is inefficient due to its low aspect ratio (except at supersonic speeds where the lift-related lift is tiny fraction of the overall drag).

@JanHudec How does this apply to high Mach number operations where most of these modern business jets cruise (Mach 0.8 to 0.9)? Wouldn’t it become more efficient in the transsonic region?

@Pugz, no. The aircraft cruise just below their drag divergence speed, that is before significant effect of wave drag, so the subsonic relation between induced and form drag applies. And note that at the altitude, the cruise speed is just something like 250–270 knots indicated and calls for rather high aspect ratio (usually over 9).


text:
What is a tactical penetration descent?
I was flying the other day by a military base and heard a military aircraft (I'm not sure what type it was) request a tactical penetration descent from the controller at a civilian airport. What is a tactical penetration descent?


text:
How critical is it if a pebble blows into a propeller?
How critical is it if a pebble blows into a propeller while taxiing and leaves a small abrasion in the leading edge? It wasn't noticed during flight and there was no apparent change in performance of the airplane, it was only noticed after parking. This is on a single engine piston aircraft.

This sort of FOD is often not an immediate problem, however introducing any slight knick or gouge to a propeller could be the beginning of a stress fracture. The last thing you want is a catastrophic prop failure whilst flying!
The correct course of action, much like any minor damage to the airframe, is to note it in the logs, have a mechanic look at it, and make the call if a repair is necessary.

What sort of repair options are typically available? Can individual blades be repaired? A weld bead deposited  followed by careful machining? The original blades are they typically cast / forged or machined into shape?

@curious_cat I'm not sure i'm qualified to answer that, however I suppose it would depend a lot on the type of propeller, specifically what it is made out of. Carbon fibre can be patched, wood can have an extra layer lamenated on top, metals may be harder I don't know. This prop overhaul company has a video of them repairing a prop

For completeness...  Foreign Object Damage (FOD).

Based on the description provided; it will have little aerodynamic (thrust, drag etc) effects, which is negligible and will not notice much in terms of performance.

small abrasion in the leading edge

After parking, maintenance guy has to check it up and take decision based on the impact, maximum allowed damage in propeller, repair availability in the current airport etc...

In case of any hesitations you should refer to the documentation of your propeller manufacturer. They contain criteria for defects, which ones can be ignored, which ones can be fixed by the aircraft owner, and which ones must be fixed by the manufacturer.
Any other action (or inaction) may cause damage to the prop, engine, and consequentially to the aircraft (be it from a crash land, or a glide land).


text:
How can one achieve supersonic cruise capabilities without afterburners?
I have read that in order to have stealth capabilities of the F22, it was designed such that it could cruise at supersonic speeds without using afterburners, which increase RCS. How is this possible? what is the technology behind this?

the F-22 is not the first aircraft that can reach supersonic speed without afterburners: the English Electric Lightning could do that in the 1950s.

A normal jet engine has subsonic flow throughout. This avoids shock problems such as choking or excessive acoustic noise.
A conventional afterburner accelerates the exhaust flow to supersonic speed, allowing it to push the plane past the speed of sound. It is placed after the turbines in order to avoid those shock problems.
A supercruise engine must deliver supersonic exhaust flow without using an afterburner. Luckily, in a hot gas the speed of sound is much higher than in a cold gas. So as long as the exhaust is not cooled before exit, its high exit velocity stays subsonic within the engine and only "looks" supersonic relative to the outside air.
The problem with this is that you cannot obtain the usual benefits of mixing bypass air with it to increase its mass flow and thermodynamic efficiency, because that would cool it and create a standing shock wave across it, slowing it to subsonic speeds.
The end result is a bit of a design compromise; relatively high mass flow through the core, relatively low bypass ratio, some sacrifice of efficiency.

1  - a more complete answer than mine

It's not that unusual, it's called Supercruise

Supercruise is sustained supersonic flight of a supersonic aircraft
with a useful cargo, passenger, or weapons load performed efficiently,
which typically precludes the use of highly inefficient afterburners
or "reheat".
[snip]
One of the best known examples of an aircraft capable of supercruise
was Concorde. Due to its long service as a commercial airliner,
Concorde holds the record for the most time spent in supercruise; more
than all other aircraft combined

The F-22 is powered by the Pratt & Whitney F119

The requirement for the ATF to supercruise, or fly supersonic without
afterburners, results in a very low bypass ratio of 0.3 for the
F119-PW-100 in order to achieve high specific thrust.

There's nothing magical about the speed of sound. Sure, there are some shockwaves that form around that speed that increase the air resistance, but getting past those is a (conceptually) simple matter of designing the plane to minimize the shockwaves, and then having enough engine power to overcome what's left.

Engine power is not enough. The engine has to achieve supersonic exhaust flow, and that is a whole different ballgame.

@GuyInchbald In order to generate thrust, the exhaust must be faster than the airplane. So, yes, to get to Mach 1, the exhaust must be faster than Mach 1. But that's true of any speed, there's still nothing magical about the speed of sound.

HiddenWindshield A difficult technical challenge and magic are not the same thing. See my main answer.


text:
Can you become FAA/TC commercial pilot with metal implants in hand?
I have a friend who wants to be a pilot and he has metal implant in his hand. It doesn’t affect his hand movement at all. I know that he will need to check it out with a Civil Aviation medical examiner first before it can certainly be okay. The problem is he is going to have to travel to Canada or USA to study aviation and do his medical. He just wants to know if in general metal implants are a no go in aviation before he travels to start his process of becoming a pilot.
( yes it’s not a question for me it’s really for a friend... I am already a pilot)

I have never had a doctor pass a metal detector over my body during a flight physical, nor have any of the health history questions asked about metal implants.  Just sayin...

There are two constants in aviation medicine. #1 Almost anything and everything could cause you to lose or be denied a medical certificate. #2 Almost anything and everything can be waived given enough time, money, and determination. Google Sgt. First Class (Ret.) Dana Bowman, a double amputee commercial helicopter pilot and CFI.

The simplest way to answer this is to have your friend call an AME in the city they intend on training and ask them if they feel they will pass the required medical for a CPL. AME's are pretty forthcoming about this kind of stuff and they will often give you an idea over the phone if they feel there is a real issue. AME's are also the ONLY people that can definitively answer this question for your friend as they are the only people that can issue medical certificates.
Generally if the desired applicant has no practically limiting factors (in this case full use of their hand) the worst I would foresee needing is a Statement of Demonstrated Ability (SODA) which basically means they can show that they can execute the required actions of a pilot.
Under the Musculoskeletal section of the AME guidelines it states:

(b) No other organic, functional, or structural disease, defect, or
limitation that the Federal Air Surgeon, based on the case history and
appropriate, qualified medical judgment relating to the condition
involved finds

(1) Makes the person unable to safely perform the
duties or exercise the  privileges of the airman certificate applied
for or held; or

(2) May reasonably be expected, for the maximum
duration of the airman medical certificate applied for or held, to
make the person unable to perform those duties or exercise those
privileges.

If your friend can perform the duties of a pilot he should be ok.

There are no AME's who post here (as far as we know... maybe one posts incognito, but that's no help either).
The best you're likely to get is, "I know a guy who..." and whatever his result was. Will that match what happens with your friend? Who knows.
The question is different, but the answer is the same as on this thread. (I hate closing "as duplicate" questions that are different than the 'duplicate,' even if the answer is the same.)
For what it's worth, pilots routinely continue to fly after hip replacement, so by itself, having metal inside you isn't remotely a big deal.
Good luck to your friend!  (And, I am serious... and don't call me Shirley.)

Better a comment than answer...

"Pilots continue to fly after hip replacement" sounds like an answer to me.

There is even one famous pilot/CFI double amputee. Sgt. First Class (Ret.) Dana Bowman.

I ment to up vote this one. Oops. My bad.


text:
Can the lift equation be used for the Ingenuity Mars Helicopter?
I am currently working on a basic simulation of the Mars Helicopter Ingenuity, and I am having trouble interpreting the outputs I am receiving from the lift equation. The equation reads as follows:
L = Cl * A * .5 * r * V²
L is lift in newtons, Cl is the coefficient of lift, A is the area of the circle traced by a blade in one rotation, r is atmospheric density, and V is the velocity of the blade. Ingenuity has two rotors 1.2 meters across, each with two blades. For this example, let's assume a rotation rate of 2400 rpm. The coefficient of lift is 0.4, the average for small helicopters. Atmospheric density on Mars is 0.020 kg/m^3. The velocity is taken 1/2 way down the blade and is found by using ((1.2/2)π * rpm / 60). Plugging in the values for Ingenuity gives:
102.871 = 0.4 * π(0.6)² *.5 * 0.020  * 5684.858 * 4
The 4 at the end is added because Ingenuity has four blades, each generating its own lift in accordance with the lift equation.
Dividing the lift by the mass of Ingenuity (1.8 kilograms) gives an upward acceleration of 57.15 m/s². Unfortunately, this number is too large to be accurate. Martian gravity is only 3.711 m/s², and whatever drag that will be generated by the thin atmosphere is likely to be insignificant in relation to such a large upwards acceleration.
So, the question is, how should lift be calculated in this case? Am I using the lift equation incorrectly, or is it simply not applicable to small drones like Ingenuity? If you have any resources for calculating lift more accurately in this case, please share them.
Thank you.

Welcome to AV Stack Exchange. I'm not familiar with the martian atmosphere, but that velocity looks like you're going much faster than the speed of sound. At that point, the lift calculation becomes different. I think it would also be useful to calculate the Knudsen number for this phase of flight (https://en.wikipedia.org/wiki/Knudsen_number) to see if continuum mechanics is applicable.

@StuartBuckingham actually, the ends of the rotor blades only reach ~0.7 times the speed of sound, and the spacecraft is never planned to exceed 3 m/s vertically. Although, if the output of my equation was correct, Ingenuity would reach the speed of sound very quickly!

THERE IS A BIT OF MISUNDERSTANDING HERE, THAT EQUATION CANNOT BE DIRECTLY USED FOR HELICOPTERS, wherever they fly
The correct equation for rotary-wing is:
$T = ½ \rho (\omega R)^2 \pi R^2 C_T$
where $\omega$ is the rotating speed of the rotor and $C_T$ is the thrust coefficient ("thrust" since rotors generate no lift) which cannot be more than some 0.03 for very aerodynamically optimised rotors. Please have a look at my answer for its correct application.

I'm not sure, but maybe this should be moved from aviation to physics?

@JosephDoggie very very hopefully not. It has 4 good answers and numerous useful comments. The subject is if far greater interest to this community than to Physics where it may well get minimal attention. It's 2+ years old .

@RussellMcMahon -- I understand.  Still, there does seem to be a lot of physics equations there, as opposed to, say, procedures for checking aircraft before taking off.

The formula is correct, but you're measuring the wrong area.
A is the area of the wing, i.e., rotor blade, not the area of the rotor disc.
When π(0.6 m)² is replaced with the correct value, about 0.6 m * 0.05 m, then the final number for lift drops by about 40 times, as does the upward acceleration, to a more plausible 1.4 m/s².
This error is also committed by the first google hit I got for lift equation rotor.
The next step to improve accuracy would be to not ignore that the root of the blade moves much more slowly than the tip, and conversely that (for Ingenuity at least) the root has greater chord and greater angle of attack.  But the math for that is much more involved: see sophit's or xxavier's answers, which are fancier than what the OP needed, and which lack this answer's "Add citations from reputable sources" post notice.

What a great response! It seems like the area misconception resulted from a mix-up between different sets of equations. This source appears to show that the full disk area is used in the more complicated airflow-based Navier-Stokes Equations. Perhaps the author of the article you linked confused this with the area for the lift equation. Also, it seems like you have answered many questions like mine on this forum. Helpful folks like you make the internet a much more useful tool, and keep communities like this one alive, so thanks for your work.

Heh, answering such questions is what keeps me sharp!

Isn't $C_L$ and $A$ dependent of each other? ie. if you choose to use area of entire rotor, the lift coefficient also describes entire rotor. For approximation like this that might be more suitable approach as you cannot add blades indefinitely to increase lift of the rotor.

Yes, the OP's formula ignores interactions between blades.  But ignoring how many blades each rotor has may matter less than ignoring the inter-blade effect of two coaxial rotors.

Why did you do $0.6 \times 0.05$? is 0.05 the width of the blade?

Yes, 0.6 m approximates the blade's span, and 0.05 m approximates the blade's chord.

THERE IS A BIT OF MISUNDERSTANDING HERE, THAT EQUATION CANNOT BE DIRECTLY USED FOR HELICOPTERS, wherever they fly
The correct equation for rotary-wing is:
$T = ½ \rho (\omega R)^2 \pi R^2 C_T$
where $\omega$ is the rotating speed of the rotor and $C_T$ is the thrust coefficient ("thrust" since rotors generate no lift) which cannot be more than some 0.03 for very aerodynamically optimised rotors. Please have a look at my answer for its correct application.

@CamilleGoudeseune: thanks for mentioning my answer which is anyway not fancier but simply correct. Using the lift equation for rotary-wing applications is physically wrong and gives wrong results, that's all. Anyway also mine answer had the post notice which is gone as soon as I cited reputable sources that you can find at the end of my answer.

There is a bit of misunderstanding here, the "lift" equation cannot be directly used for helicopters, wherever they fly and whatever the chosen answer says.

The correct equation for rotors and propellers is the "thrust" equation which is formally similar to the lift equation but uses the speed due to the rotation and the disk surface instead:
$T = ½ \rho (\omega R)^2 \pi R^2 C_T$
where $\omega$ is the rotating speed of the rotor, $R$ its radius and $C_T$ is the thrust coefficient ("thrust" since rotors generate thrust, not lift) which cannot be more than some 0.03 for very aerodynamically optimised rotors.

The equation given in the question is used to calculate lift generated by a wing (or, more generally, an aerodynamic body) with surface $S$ when it moves in a fluid of density $\rho$ with a well defined speed $V$:
$L = ½ \rho V^2 S C_L$
Why is this equation not suitable for rotary-wing applications? The speed $V$ seen by an helicopter blade is basically given by the sum of 3 different terms:

the speed due to its rotation around the rotor shaft; this speed varies linearly along the bladespan from 0 at the rotor head to $\omega R$ at the tip, where $\omega$ is the rotating speed and $R$ the bladespan;
the speed at which the whole helicopter is flying;
during its rotation, the blade bumps into the wake shed by the previous blade; and after one complete rotation, it bumps into its own wake; these wakes also modify the local speed seen by the blade.

The sum of these 3 terms gives a total speed which continuously changes in space and time and that definitely cannot be represented by one single constant value $V$.
So, can the lift equation be used for an helicopter? No, because the blade does not move within a uniform and well defined airflow with steady speed $V$.
That equation can be used for airplane, cars, cyclists on bike, bridges and so on, but not for helicopter or, more generally, not for rotary-wing aircraft.
Is it possible to use an equation as easy as that one for rotary-wing applications? Yes it is, but in the helicopter world nomenclature is a bit different: lift, per definition, is the component of the aerodynamic force perpendicular to the airflow. As discussed, the airflow seen by a rotor is the sum of 1.+2.+3. and it is definitely not constant, it changes continuously. Therefore defining a lift for a rotor is quite difficult simply because the airflow (and it's perpendicular component) changes continuously. Therefore for rotor not lift rather thrust is defined: it is the aerodynamic force parallel to the rotor shaft and positive upward.
Analogously to the lift equation, also the thrust can be represented as the product of a density, a velocity (squared) and a surface but with the following distinctions:

as a velocity, the velocity of the rotating blade tip is used, that is $\omega R$
as a surface, the area $A$ of the rotor disk is used, that is $\pi R^2$

Therefore for rotary-wing application, the following thrust equation is used:
$T = ½ \rho (\omega R)^2 \pi R^2 C_T$
(Note: in USA the $½$ is normally dropped)
$C_T$, a bit like $C_L$, depends on several geometrical factors and trade-off, but in general its maximum value is within the range 0.008 and 0.028, according to how much optimised the rotor is.

Intermission
Let's do the math for Ingenuity making the following assumptions

$\rho = 0.01 kg/m^3$ Note: density on Mars changes according to season from 0.01 to 0.02; we use the lowest value in order to be conservative;
$\omega$ = 2400$rpm$ = 250$rad/s$
$C_T$, we suppose that the rotors are of a very optimised design = 0.028.

Then we get, remembering that Ingenuity has 2 rotors:
$T =2 \cdot ½ \cdot0.01(250\cdot0.6)^2 \pi 0.6^2 \cdot 0.028 = 7.125N$
Since $g_{mars} = 3.72m/s^2$ that corresponds to 2.17kg thrust, which is enough to hover and manoeuvre Ingenuity with its 1.8kg weight.

Bonus material: what if we do use the lift equation (in the correct way)?
Is, in the helicopter world, the lift equation as given in the question really useless? If the hypothesis is done that the blade is composed by a continuous sequence of adjacent airfoils, then that equation can be locally used for each of these airfoils in the sequence.
So, for each slice of blade that equation can be used to get the local lift (and the local drag); afterward the local lift (and drag) is summed up (integrated) along the bladespan, from the rotor head till the tip, to get the total lift and drag; and finally the total lift and drag is decomposed in 1) a force parallel to the rotor shaft, which is the thrust generated by the rotor; and 2) a force perpendicular to it, which gives the torque needed to make the rotor spin.
Obviously that equation can be used if it is know the local speed $V$ on each slice of blade as given by 1.+2.+3. As seen, 1. and 2. are know (since $\omega$, $R$ and the helicopter speed are known); the speed 3. due to the wake depends itself on the lift and gives rise to a vicious circle: the lift depends on the wake and the wake depends on the lift! This vicious circle is broken by either using simplified models of the wake and/or wind tunnel measurements and/or CFD simulations.
Let's break the vicious circle via a simplified model that we will later test again on Ingenuity. If the simplification is done that:

the helicopter is hovering, i.e. term 2. is null;
the wake in/upon the rotor is constant everywhere, i.e. term 3. has a constant value;
the blades have a hyperbolic twist (this is normally true for a propeller blade, helicopter blades have a more linear twist but using a hyperbolic one the math is easier);
and each airfoil is operating at its $\alpha$ of maximum efficiency i.e. maximum $C_l/C_d$;

then we get that:
$C_T=¼ \frac{N_b c_{tip}}{\pi R} C_{l_{\alpha}} \alpha_{@maxC_l/C_d}$
where:

$N_b$ is the number of blades;
$c_{tip}$ is the chord at the tip of the blade;
$C_{l_{\alpha}}$ is the slope of the airfoil's lift coefficient;
and $\alpha_{@maxC_l/C_d}$ is the airfoil's $\alpha$ for maximum efficiency.

For Ingenuity we have:

$N_b=4$;
$c_{tip}=0.06m$ (ballpark estimate)
$C_{l_{\alpha}}=2\pi$ (typical thin-airfoils value);
and $\alpha_{maxC_l/C_d}=8°$ (this value really depends on the airfoil; I have chosen a standard NACA 2412 at a very low Reynold number).
This gives:

$C_T=¼ \frac{4\cdot0.06}{\pi\cdot0.6} 2\pi \cdot0.14=0.028$ Yeah

Yet another bonus material: momentum theory
A first approximation of the thrust can be obtained also using the simple momentum theory which gives:
$ T = \sqrt[3]{2 \rho A P^2} $
Where:

$A$ is the rotor area;
$\rho$ is the density;
$P$ is the power needed to generate the thrust.

To see the simple momentum theory in action for Ingenuity, please have a look at this answer.

Useful references:

Principles of Helicopter Aerodynamics by J. Gordon Leishman
Helicopter Aerodynamics Volume II by Ray Prouty
Aerodynamics of the Helicopter by Alfred Gessow, Garry C. Myers

Excellent answer! I notice that to account for there being two rotors, you simply multiply by two.  As they are coaxial, I imagine that further complicates the vicious circle of interaction, so my first thought that multiplying by two was leaving something out. My second guess is that these effects are captured by the coefficient of thrust for the assembly differing from that of each rotor considered individually. Am I making any sense here?

@sdenham: yes you're right, two coaxial rotors normally need some 15% more power than two isolated rotors, the total thrust being the same. And for the thrust I made the indirect simplification that all the complications are incorporated in the $C_T$

This is such a great and detailed answer. I don't understand it at all, but kudos ;)

@Jamiec it's not that difficult: for airplane speed is simply V, for rotating blades it's V + rotation speed + wake speed ;)

In the USA the 1/2 is dropped?  Could it be factored into CT?  Going to try your equation for an air plane prop.

@RobertDiGiovanni: yes sure, it's factored in $C_t$. Let me know what you get.

For a first approximation, you may take 0,7R as the representative station, but the area should be the area of the blade, not that of the swept surface.
Let's take 2400 rpm. That's 251 rad/s. And let's take the airspeed V at 0,7R => 0,7 * 0,6 = 0,43  0,43 * 251 = 105 m/s
Let's assume that the blade chord is 0,1m. The blade area A would be 0,1 * 0,6 = 0,06 m2
L = Cl * A * .5 * r * V²
Takin Cl as 0,4
Lift = L = 0,4 * 0,06 * 0,5 * 0,02 * 105^2 = 2,7 N
But we have four blades, so we may take 4 * 2,7 = 10,8N as the total lift.
Gravity acceleration in Mars is 3,7 m/s2, so that ingenuity's weight will be 1,8 * 3,7 = 6,6 N
You have plenty of lift...!
We can try to be more accurate, by integrating the lift along the blades. A elementary derivation/calculation follows:
We start with the expression for lift:
$$L=\frac{1}{2}\cdot C_{L}\cdot v^{2}\cdot S\cdot \rho$$
Let the radius being r, the chord c, and the pitch p. Then, we take a rectangular element of blade surface:
$$dS=c\cdot dr$$
For that blade element, and in differential form,
$$dL=\frac{1}{2}\cdot C_{L}\cdot v^{2}\cdot c\cdot dr\cdot \rho$$
Taking v as a function of rotor angular speed omega at radius r in air of density rho and an inflow w (blowing from below the disk) $$v^{2}=\Omega ^{2}r^{2}+w^{2}$$
Taking CL as a linear function of the AoA, itself determined by the pitch p and $$\arctan (w/\Omega r)$$ then we have $$AoA=p-\arctan (w/\Omega r)$$
We may take the airfoil NACA0012. From the NACA graph, and with the AoA in radians, $$C_{L}=5,72\cdot AoA$$
We end up with $$C_{L}=5,72\cdot ((p-\arctan (w/\Omega r))$$
The angles being small, they may be taken as the tangents, in radians.
Inserting CL and v2 in $$dL=\frac{1}{2}\cdot C_{L}\cdot v^{2}\cdot c\cdot dr\cdot \rho$$
We arrive to $$dL=2,86\cdot c\cdot \rho \cdot (p\cdot \Omega ^{2}\cdot r^{2}+w\cdot \Omega r+p\cdot w^{2}-w^{3}/\Omega r)\cdot dr$$
For constant values of chord, air density, inflow, rotor angular velocity and blade pitch, the whole expression may be integrated, but if we assume that inflow is zero, and taking only the first term, that is the important one, we have that, for one blade:
$$L=2,86\cdot c\cdot \rho \cdot p\cdot \Omega ^{2} \int_{0}^{r}r^{2} dr$$
Inserting values for the Mars helicopter:
rotor radius 0,6m,
Omega = 2400 rpm = 251 rad/s rho = 0,02 kg/m3 chord = 6 cm blade pitch = 9º = 0,157 rad
$$L=2,86\cdot 0,06\cdot 0,02 \cdot 0,157\cdot \ 251 ^{2} \int_{0}^{0,6}r^{2} dr$$
Integrating…
$$L=2,86\cdot 0,06\cdot 0,02 \cdot 0,157\cdot 251 ^{2}\cdot  0,6^{3}/3$$
We end up with a lift of 2,4 newton per blade...

Awesome! I actually already have a setup for calculating lift for individual blade segments--I just wanted to keep things simple in this post.

Two ways of calculating helicopter rotor thrust: with the simplified momentum theory, and with finite element methods looking at the lift/drag equations of rotating blades.

Momentum theory. The simple way is to use momentum theory, where the reference area is the rotor disk area. The relevant dimensionless aerodynamic coefficient is $C_T$ rather than $C_L$, as in: $$T = C_T \cdot \rho \cdot A \cdot \Omega^2 \cdot R^2$$ with:
$C_T$ = thrust coefficient;
$\rho$ = air density [kg/m$^3$];
A = rotor disk area [m$^2$];
$\Omega$ = angular blade speed [rad/sec];
R = blade length [m]
The $\Omega$R in this equation is the blade tip speed. $C_T$ would be in order of magnitude of 0.004 - 0.008 in normal circumstances, depending on number of blades, twist etc. Some more information in this answer, loads more in J. Gordon Leishman, Principles of Helicopter Aerodynamics.

Blade element theory. With this method the reference area is the total blade area - we compute the lift/drag at every section of the blades, then integrate over the blade length, as per example for propellers in this answer. The helicopter blade will have:

twist: variable angle-of-attack
sometimes variable blade cord;
variable speed as a function of distance to the rotational centre and forward/sideways speed;
variable boundary layer and Reynolds effects.
a large amount of tip loss,
a bit of blade lacking around the rotor centre, so blade length integration boundary does not start from zero.

SO the method OP used was the blade element method, but with a couple of things not entirely right:

Computing lift per blade, the reference area is the blade area, not the rotor area.
Velocity should be taken as an integral over the blade length, not as a constant at half blade length (the lift generated at the tip is four times as much!)
The tip/root losses should be taken into account.

For simple birds eye view considerations, nothing beats the momentum method.


text:
Is there a reason that commercial jets do not have a “dead man’s switch”?
Following high-profile cases of disappearances such as MH370, as well as the lower-profile but no doubt common radar infringement and level / airspace busts that we don’t hear about often, is there a reason that commercial aircraft do not have a dead man’s switch?
Although I have no personal experience in aviation, this would have seemed a very simple way to answer the question of “was it pilot incapacitation?” Either it was, in which case the dead man’s switch triggers some sort of alert to ATC / on the FDR, or it was a  case of outside interference in which case either they a) fake incapacitation by triggering a dead-man’s alert, which the scrambling of jets would soon disprove or b) they don’t trigger a dead-man’s, and so at least pilot incapacitation is ruled out in the subsequent investigation.
Of course there is the question of how a dead man’s switch could be implemented, however given the number of SOPs already taught, this hardly seems a limiting factor.

On a train, a released dead-man switch brings the train to a stop, since that's presumably safer than letting it run out-of-control at high speed. What sort of safer alternative do you think exists for a flying aircraft? "Just stop here, where-ever we are" clearly isn't it!

Welcome to aviation.SE! Any system would have to be more reliable, flexible and cost-effective than the second pilot, who's already sitting there. If an incident incapacitates both pilots simultaneously then it's probably something catastrophic (or deliberate) and an automated system might not help. For single-pilot aircraft Garmin has a system that monitors the pilot and can initiate an autoland but it's expensive and requires an autothrottle, which many smaller aircraft don't have.

The OP is proposing a dead-man's switch that sends a radio or satellite message, not one that autolands the plane.

@RalphJ Yes sorry, I probably wasn't clear enough in my original question - I obviously was not proposing for a dead-man switch which somehow brings the plane out of the sky (though in the future with Garmin auto land for example, this may be a possibility). I in fact meant that the activation (or deactivation, I suppose?) of the dead-man switch would at least alert the relevant authorities that something is amiss. Sure, there's still going to be an accident, but at least you know that the pilots were, for whatever reason, not in control of that aircraft.

One good reason why not is because somebody will have to hold it the entire time.  Pilot’s have other things to do too.  And where do you stop?  Why not require them on every car on the highway?

@MichaelHall on trains the switch needs to be operated from time to time, not continuosly. Pressing it continuosly would have the same effect as not touching it at all.

@orique, I have never driven a train, but that isn’t my understanding of what a dead man switch is. In the Navy it was a spring loaded switch held closed. If the holder perished hand would relax, causing the spring to open the switch. What you are describing sounds more like the occasional “I’m still alive and don’t need any help” button like they use in nursing homes.  I think the needs of trains might be different given that they likely don’t maintain constant radio contact.

@MichaelHall at least in Spain the "still alive" approach is used, don't know about other countries

@MichaelHall as far as I am aware a train's dead man switch needs to be 'reactivated' every so often, following an audible cue. This is more what I imagined: in the same way that pilot's must regularly report their position when flying transatlantic, this switch would require them to regularly register their attentiveness. Of course, the constant radio contact does mitigate somewhat the need for this.

@Harvey, yes, I would agree that required position updates via radio renders such a button operated check-in moot.

@Harvey - Plus, the radioed position report provides confirmation that the pilot is alert, responsive, and functional. A simple push of a button would only confirm that the pilot is alive. Even if it were a specific code the pilot must enter. YouTube videos of ATC communications with pilots suffering from hypoxia. A pilot might be able to operate an autopilot or a coded deadman’s switch through rote or muscle memory while not being usefully conscious.

@Harvey I’ve ridden in a modern train cab; the alerter only goes off after X seconds with no control inputs, and if they still don’t do something (like press the reset button) within another X seconds, it dumps the brakes. This system was developed after a few crashes where it was discovered drivers were just putting a brick on the older spring style deadman switches to prevent false positives.

Very intriguing question. I do wonder what it actually solves, though. Does it only solve the mystery of whether the very rare instance of incapacitation of the flight crew occurred? The Cockpit Voice Recorder does that if it is recovered. In the rare event that the CVR is unrecoverable, the dead man’s switch would only answer if pilot input occurred prior to a crash. How would it save lives? The pilot should already be in periodic verbal communication with Air Traffic Control.
On the other hand, a global monitoring system of the aircraft would make more sense. Each commercial aircraft already has transponders. In the US, each aircraft has ADS-b. This gives a nearly real-time position of each aircraft that can be tracked and archived. This would give searchers a better idea of where to search for survivors. Incorporating a 406 MHz PLB-style beacon into the Flight Data Recorder and the CVR would be even more advantageous in solving the mystery of what happened. Even an automated system that alerts ATC of when contact is lost with the transponder or ADS-b, or when the CVR or FDR 406 beacons were activated, or when the aircraft deviated too far off its filed flight plan, would be better than a dead man’s switch.
After all, in today’s era of modern telemetry and aviation security, would scrambling fighters be of any benefit to the passengers and crew on board while the aircraft is in flight? They would have no way of intervening to prevent a crash.
It’s only benefit would be to alert ATC of a terrorist highjacking in the event that the target were close to the airport. And, the overtaking of the cockpit happened in the last phases of flight. Now, the security of modern cockpits and heightened alertness makes the notification of ATC of a more likely.

Great answer, and I couldn't agree more that tracking of the FDR and CVR would be the  best solution. I wasn't aware of all these other measures that could be / are already in place to keep tabs on aircraft. Still, for a sector so fond of redundancy, and of leaving nothing to chance, it does still shock me that a more robust solution other than "contact ATC, and if you don't you're probably dead anyway" does not yet exist.

@Harvey - That brings us back to my original question. What would this solve? How does a deadman’s switch make aviation safer? If it does not make things safer, it is not a redundancy measure. A Cockpit Video Recorder would be a redundant system to the FDR & CVR. A deadman’s switch would only make sense if remote control of the plane were implemented as well. Also, aircraft flying IFR have mandatory reporting points and procedures. Especially in areas where radar coverage is sparse. Contact with ATC should be fairly regular and constant.

@Harvey - The system to which the mandatory reporting points is redundant is the fact that it is a requirement of all aircraft flying IFR in all airspace (this includes all commercial airliners flown by scheduled carriers), all aircraft flying VFR in 4 of the 6 types of airspaces, and all aircraft flying within 4 miles of an airfield with operating ATC at or below 2500 feet AGL to establish and maintain two-way communication with ATC. If 2-way communication can not be established and maintained, specific Lost-Comms procedures must be followed.

Seems that a tear-away floating PLB for water 'landings' would be a better plan. If a general vicinity is signaled, the chance of finding the FDR and CVR before they stop pinging become much greater. Non-water 'landings' are easier to find.


text:
Does the engine mount of a turboprop engine in an aircraft experience engine torque?
When the turboprop engine is active, the propeller shaft rotates and causes the propeller to rotate as well, which in turn generates thrust. The propeller shaft rotates because of the torque (or twisting moment) created by the turbines of the engine and then which is conveyed to the propeller shaft. My question is that if I am trying to conduct a FEA analysis for the mount at which this engine is attached, should I also apply this torque on it or not? I was thinking that I shouldn't apply any kind of torque at this engine mount since I don't think that engine itself is rotating at all because I don't think there is a physical connection between the turbines (or propeller shaft) and the engine casing (to which the mount is attached). But it is recommended that the torque should always be taken into account while conducting such analysis. Moreover, I don't know the torque (if) experienced by the engine casing and mount would be equal to what the propeller shaft or turbine is experiencing.

Before you run your FEA, draw a free body diagram.

@ZeissIkon, the free body diagram can be drawn. That is what my question is exactly; I am concerned about the engine mount only and I don't know if I am supposed to input torque on it or not. As far as free body diagram for engine is concerned (although which is not my purpose of study) I cannot include the airflow near the turbines, which I think is experiencing the torque (but in opposite direction) in regards with Newton's third law of motion.

I can't tell you for sure, but a helicopter's main rotor causes a torque on the entire aircraft, that is counteracted with the tail rotor. I imagine the propeller itself on a turboprop causes a torque on the mount.

Draw the free body diagram for the engine -- what's opposing the torque the engine applies to the propeller?

@ZeissIkon, since the turbine applies the torque on the propeller shaft, there must be something which should oppose this torque. The turbine doesn't have a direct physical connection with the engine casing. So I am thinking its the airflow around the turbines which is opposing this torque.

@ZeissIkon, just for the record, I don't necessarily believe that there should be something to oppose the torque on the propeller shaft. I mean it can undergo Newton's second law instead of third.

@RameezUlHaq A torque on the shaft can't produce an acceleration, unless that means the aircraft spinning out of control.

If the torque reaction isn’t applied through the engine mounts then where is it applied?

@stevederekson555, I was talking about rotational acceleration on the propeller shaft only and hence on the propeller, not on the engine or engine mount or the aircraft itself.

@RameezUlHaq I think that a propellor always transmits a torque to the shaft.

@stevederekson555, everybody knows that here, dude. Actually, its the other way round. The shaft transmits the torque to the propeller. But this idea that torque results in an opposite torque on the engine casing and hence on the engine mount (since mount is connected to the engine casing) confuses me.

@RameezUlHaq You misunderstood my comment. In a stationary state the aerodynamic torque on the propellor that is transmitted to the shaft is greater than what is transmitted to the propellor, so the shaft responds with a reaction torque which is what causes stress on the structure.

@stevederekson555, stress on what structure? Stresses on the turbine blades only, or the stresses on the engine casing itself (which is attached to the mount)?

The torque applied to the propeller originates in the turbine section, from the lift forces being applied to the turbine blades, being little wings going round and round, by the accelerating gases leaving the burner can.
Therefore, it's the lift forces of the air mass passing the blades that are generating a Newtonian reaction force path from gas pressures acting in the opposite direction through the stator vanes and engine case.  So the torque generated at the center of the turbine disc, where the through shaft connects, is seeing a reaction torque force in the opposite direction being applied to the engine casing, wanting to rotate the engine in the opposite direction along the turbine shaft axis.
Forget about the propeller for a minute and imagine you have a turboshaft engine in a helicopter, where there is just a drive shaft running to a completely separate transmission unit. The engine doesn't care that there is a rotor on top of the transmission, it just knows that it's spinning a shaft against resistance. The torque driving the rotor transmission is still originating in the engine's turbine section, and wanting to rotate the engine case in the opposite direction, restrained by the engine's mounts.  Of course the torque forces are also generating a reaction force within the transmission wanting to rotate the helicopter opposite to the rotor's torque, necessitating a tail rotor.
With a turboprop engine with an integral propeller reduction gearbox, this reaction torque is being absorbed by the engine mounts, along with thrust loads and gyroscopic precession forces from the propeller, transferred to the propeller gearbox through the shaft bearings of the prop, to the engine case, to the engine mounts.
The result is that the load at any particular mount is the sum of the forces acting at any given time at that point from thrust, torque and precession forces acting on the case.  On the other hand, a turboshaft-based turboprop like the General Electric T-64 has the propeller gearbox on a separate mount, so that engine mounts only see rotational torque from the turbine section.

I'm hearing you on the point a transmission or reduction gear can take the prop (or rotor) load away from the center of the drive shaft, producing torque.  With a helicopter, it is geared to drive the rotor that is 90 degrees from the jet turbine shaft, no?

John K, just to be clear here, so the torque that the propeller shaft is experiencing is exactly equal to what the engine casing is experiencing, and they both are equal to the torque produced at the center of the turbine disc? Or the torque produced at the center of the turbine disc is equally distributed to the propeller shaft and the engine casing?

@RameezUlHaq: To a first approximation, the shaft and the casing both experience the full torque produced by the disc.  It’s like if you have a vertical stack of cardboard boxes with a heavy weight on top: each box experiences the full weight, the weight doesn’t get distributed through the column.

It will depend on where the vanes are.
Turbines have vanes which redirect the flow, on the inlet side helping avoid compressor stall, and on the exhaust side capturing the rotational energy and converting it to thrust.
These vanes are rigidly attached and so the air hitting them will cause a rotational torque on whatever they are attached to.

Good point.  Torque Force on the stationary vanes will try to twist the nacelle.  So twist a tube made of aluminum foil and one made of 1/4 inch steel.  What happens?  So again, is there torque on the pylon holding the nacelle?.

Sigh, always these downvoters who never explain. I'd love to know what part I have missed in the very simple Newtonian mechanics of air impinging on a stationary surface.

Yes, it does...
We can regard the turbine as a 'black box' whose output is the rotation of a shaft. If that output finds a resisting torque, of any sort, the entire 'black box' will react, on its mount, with a torque of equal magnitude and opposite sense.

This is a great question because it challenges generally accepted principles of prop torque generated by a piston engine.  One clue in the question states "a twisting moment creating by the turbine", and there were thoughts of the slender pylons on jets: is there a torque stress on the motor mounts?.
Certainly a twisting stress on the turbine shift, whether it is run through a transmission or not.  So let's load the prop while the jet is running.  Notice that blades are symmetrically arranged around the shaft compared with pistons pushing one at a time away from the center of rotation.
The symmetrical push on all the turbine blades by the jet exhaust gasses does not produce a torque force on the mounts, only a torsional stress on the shaft.  The symmetrical drag load of the prop blades does not produce a torque force on the mounts, only a torsional stress on the shaft.
One may surmise that if the turbine torque and the prop load torque are balanced around the center of rotation, there is no torque on the motor mount.
However, if one component comes out of balance (such as the prop), the motor mount may be easily torn off.
For more on turbo-prop designs, check out this question.  Seems that Pratt and Whitney designers had some thoughts on the torque issue too.
And in this report the "ovalization" of the nacelle seemed to be of greater concern, solved by increasing the pylon attachment points from 1 to 2, spaced 120 degrees apart.

The mounts may experience torque when one changes the direction the spinning works is traveling (gyroscopic precession).

@Rameez Ul Haaq In this paper the torsional moment is drawn at the aft section of pylon. See page no. 52. Maybe this would help.

@Auberron thanks, yes, we must also consider torque on the mounts from thrust.  Notice the pylon attachment points to the wing  would also be stressed in that manner.

Your answer is correct up to the penultimate paragraph. However, you will find that turbine torque is not balanced with prop torque - by design(!) For the engine mounts not to experience torque, the propwash and exhaust angular momentum most be equal and opposite. That would mean a lot of angular momentum is wasted in the exhaust stream.

@Sanchises by definition a "bearing" holds a moving object in place.  Ball bearings allow an object to rotate while it is under gravitation force.  In a jet the thrust is backwards, so the bearing must prevent the turbine from being pushed out the back of the nacelle (as well as supporting it under gravity).  Picture an astronaut in space trying to turn a screw (without support the astronaut turns).  So one could have exhaust gasses from a rocket attached to a turbine to rotate the astronaut the other way (creating twisting or torsion).  The thrust net moves everything forward!

So, if we want to turn a screw a larger exhaust turbine is needed, so more energy is extracted and less action/reaction thrust is produced (why turbo props jets get most thrust from the prop) (and I'm starting to wonder about fans too).

Net thrust is forward, of course. I'm saying that there's typically guide vanes in place to remove angular momentum from the exhaust (also, intermediary turbine stators) that transfer a net torque to the engine housing. Bearings play a negligible role in the transfer of angular momentum (unless friction is very high).

@Sanchises yes, I agree, also with Kenn Sebesta on that point.  And you are right with net thrust forward, the bearings must prevent the prop (or fan) shaft from being pulled out of the nacelle too.  So the loads on those bearings are fore and aft.  The twisting torque force (of the vanes) seems to be taken up by the rim of the nacelle.  I remember steel beer cans (compared to thin aluminum).  Still not sure if that twist puts any load on the pylon.  Great discussion anyways, thanks.

@RobertDiGiovanni, the twisting on the fixed vanes (also known as stators) is counter acting the torque produced at the rotors of the turbine (which in turn rotates the propeller shaft)? If it does, then it makes sense that why the engine mount would be taking the same amount of torque produced by the turbine (or to be precise, produced by the rotors of the turbines). And also, if what I said is true, then would the engine casing start rotating in the opposite direction to the shaft rotation if not mounted?

I'm saying that any engineer worth their salt will make sure that there is a torque load on the pylon because else they're wasting valuable kinetic energy on angular momentum in the exhaust.

@RameezUlHaq the greater danger seems to be deformation of the nacelle.

Interesting discussion.
My first thought is that a pure turbojet engine applies zero torque to the mountings but with a turboprop the mountings feel all the propeller torque. Not to be overlooked are the 1P and gyroscopic loads from the propeller (and gyros from engine of course) which are significant. Stressing the mounting for a turboprop is rather more complex than a pure jet !
And I'll add my first stressing wisdom - draw a free bogy diagram!

Why would the propeller torque be experienced by the mount if there is no 'one to one' physical connection between the propeller and the engine casing? How can one calculate the gyroscopic loads from the propeller?

There is physical contact between propeller & engine casing via the gearbox.

Gyro loads are calculated using the rotational mass inertia, revs and pitch & yaw rates.


text:
How can the flap deflect to higher angles (than the wing) without experiencing loss of lift?
For a simple aircraft wing, assume the stall angle of attack of this wing is 16 degrees. Now, if I use the same airfoil as used in the wing for the flaps, I am able to deflect the flap at quite greater angles. How is that possible? If I want to CFD analyze the flap geometry separately, can I use the maximum deflection of the flap as the angle of attack of incoming air in a CFD analysis? [Meaning will the CFD results obtained by isolating the flaps from the wings and conducting the analysis at an angle of attack the same as the maximum possible deflection of the flap, will be the same as when I conduct the analysis of the flaps together with the wing]?
How can I determine the maximum deflection of the flap? How to choose the best flap configuration for an aircraft? How determine the maximum possible deflection for the ailerons as well?

No, you can't analyse the flap geometry separately. It is not a separate element, it is moving part of the wing that allows changing its camber. And when you consider the actual chord as the leading edge moves with the slats and the trailing edge moves with the flaps, the critical angle of attack does not change much.
(I don't have much idea how to actually design a flap, so leaving this as comment only).

Would this answer help?

And even if you wanted to "analyze" the flap separately, the relative wind is altered by the wing it is attached to.